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The question we were given was (where $^nC_c$ is $n$ choose $c$):

Show, using induction and the fact that $^nC_c + ^nC_{(c+1)} = ~^{(n+1)}C_{(c+1)}$, the "hockey stick theorem": the sum from $k=c$ to $n$ of $^kC_c$ $=~^{(n+1)}C_{(c+1)}$ for all appropriate values of $n$ and $c$.

I had originally thought I got the answer but my professor mentioned in class that $P(0)$ isn't the base case and now I'm completely confused...

For the base case I plugged in $0$ as $n$, and then $0$ as $c$ since $c=0$ is the only appropriate $c$ when $n=0$, and then showed that $^0C_0 =~^1C_1.$

Could someone explain where I went wrong... I'm not sure what other base case there is.

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  • $\begingroup$ If you define $\binom{n}{k} = 0$ when $k > n$, you could start with $n = c = 0$. I suspect your professor considers appropriate values of $n$ and $c$ to be nonnegative integers $n$ and $c$, with $n \geq c + 1$. $\endgroup$
    – N. F. Taussig
    May 21, 2015 at 19:40

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